A rectangular plot, 18 ft by 19 ft, is to be used for a garden. It is decided to put a pavement inside the entire border so that 110 square feet of the plot is left for flowers. How wide should the pavement be?
width of pavement --- x ft
original plot has area of 342 ft^2
inside part is 18-2x by 19-2x, where x > 9
(18-2x)(19-2x) = 110
342 - 74x + 4x^2 = 110
4x^2 - 74x + 232 = 0
2x^2 - 37x + 116 = 0
(x-4)(2x - 29) = 0
x = 4 or x = 29/2 = 14.5
the pavement should be 4 ft wide
check:
flower bed is 10 by 11 or 110 ft^2
To determine the width of the pavement, we need to subtract the area left for flowers from the total area of the plot.
1. Calculate the total area of the plot:
Area = Length x Width
= 18 ft x 19 ft
= 342 sq ft
2. Subtract the area left for flowers from the total area:
Area of pavement = Total area of the plot - Area left for flowers
= 342 sq ft - 110 sq ft
= 232 sq ft
3. Assuming the pavement is of the same width on all sides, we can consider it as a rectangular strip.
Let's say the width of the pavement is 'x' ft.
4. The dimensions of the inner garden (excluding the pavement) would be:
Length = Length of the plot - 2x
= 18 ft - 2x
Width = Width of the plot - 2x
= 19 ft - 2x
5. Calculate the area of the inner garden:
Area of inner garden = Length x Width
= (18 ft - 2x)(19 ft - 2x)
6. Set up an equation to solve for 'x' using the area of the pavement and the area of the inner garden:
Area of pavement = 232 sq ft
Area of inner garden = (18 ft - 2x)(19 ft - 2x) sq ft
(18 ft - 2x)(19 ft - 2x) = 232 sq ft
7. Solve the equation to find 'x'.
Simplifying the equation, we get:
4x^2 - 74x + 306 = 0
This equation can be solved using factoring or the quadratic formula. Factoring is not applicable here, so we will use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a = 4, b = -74, and c = 306.
Evaluating using the quadratic formula, we find two values for 'x': x ≈ 3.40 ft and x ≈ 13.60 ft
8. Since the width of the pavement cannot be negative, we can disregard the negative value.
Therefore, the width of the pavement should be approximately 3.40 ft.
So, the pavement should be approximately 3.40 ft wide.
To find out the width of the pavement, we need to determine the area of the plot without the pavement, and then subtract it from the total area of the plot.
First, let's calculate the area of the rectangular plot: length x width = 18 ft x 19 ft = 342 square feet.
Next, we subtract the remaining area for flowers from the total area of the plot to find the area of the pavement: 342 square feet - 110 square feet = 232 square feet.
Since the pavement surrounds the entire border of the plot, it will have a uniform width on all four sides. Let's assume the width of the pavement is "w" feet.
To calculate the area of the pavement, we subtract the area of the plot without the pavement from the area of the plot with the pavement: (18 ft + 2w) x (19 ft + 2w) - 18 ft x 19 ft = 232 square feet.
Expanding the equation: (18 ft + 2w)(19 ft + 2w) - 342 square feet = 232 square feet.
Multiplying out the expressions: (342 +36w + 38w + 4w^2) - 342 square feet = 232 square feet.
Simplifying the equation: 4w^2 + 74w = 232 square feet.
Now, we solve the quadratic equation for "w" by setting the equation equal to zero: 4w^2 + 74w - 232 = 0.
To solve this quadratic equation, we can factor it or use the quadratic formula. Factoring in this case is more difficult, so we'll use the quadratic formula:
The quadratic formula is: w = (-b ± √(b^2 - 4ac)) / (2a).
For our equation, a = 4, b = 74, and c = -232.
Plugging in these values, we get: w = (-74 ± √(74^2 - 4(4)(-232))) / (2(4)).
Simplifying further: w = (-74 ± √(5476 + 3712)) / 8.
Combining like terms: w = (-74 ± √9188) / 8.
Using a calculator or math software, we find that √9188 ≈ 95.88.
Substituting the value: w = (-74 ± 95.88) / 8.
There are two possible solutions for "w":
1. w = (-74 + 95.88) / 8 = 21.88 / 8 ≈ 2.74 ft.
2. w = (-74 - 95.88) / 8 = -169.88 / 8 ≈ -21.24 ft.
Since the width cannot be negative, we discard the second solution.
Therefore, the width of the pavement should be approximately 2.74 feet.